 In this video, we provide the solution to question number 12 for the practice exam number two for math 1060, in which case we were asked to prove the trigonometric identity, cosine theta over one plus sine theta is equal to one minus sine theta over cosine theta. Both sides seem about as complicated. I will confess though that sums and differences in the denominator is much worse than having in the numerator. So I'm actually gonna start with the left hand side and we wanna prove this. So the left hand side is equal to cosine of theta over one plus sine theta. So I always recommend that just start with LHS for left hand side and then write what it is verbatim. Then what we're gonna do next depends on what we wanna do. Notice that we want to get rid of the plus in the denominator and we wanna have a minus on the top. And notice of course that the denominator of the left hand side is the conjugate of the numerator of the right hand side. So this suggests to me that maybe I can use conjugates to help me out here by times top and bottom by one minus sine theta. This is a common trick we've seen here that using conjugates can often suggest that a Pythagorean identity will come into play here. So in the numerator, I'm gonna leave it factored for a little bit. Just for a moment to see what happens. Cosine theta times one minus sine theta. The thing is I'm gonna multiply out the denominator in this situation because that's the conjugate in play here, one plus sine theta times one minus sine theta. If you foil that out, you're gonna get a one. You're gonna get a minus sine but a plus sine they cancel out and you're gonna get minus sine squared theta right there. The next thing we're gonna get is that we're going to get a cosine squared by the Pythagorean equation. Remember that cosine squared plus sine squared is equal to one. So if you solve for cosine squared, you're gonna get that cosine squared theta is equal to one minus sine squared. So a modification of the Pythagorean identity does come into play right here. We're using that to get that cosine theta times one minus sine theta is equal, or over, excuse me, cosine squared, like so. And so then this shows us the benefit of delaying factorization until it's really necessary. I like to call this lazy computation. It seems like it's a lazy thing to do. It's actually efficient though. We're waiting to compute things until we actually need them. So I didn't multiply out top because there was no need to do so yet. And that actually helps me out here. The cosine on top cancels was one of the cosines on the bottom. And so we end up with one minus sine theta over cosine theta, which you'll notice, this is now the right-hand side. So as we went from the left-hand side to the right-hand side, that then completes the proof. And now we have proven this trigonometric identity.